Download Thm 27.1: Let X by a simply ordered set having the least upper

Thm 27.1: Let X by a simply ordered set having the
least upper bound property. In the order topology,
each closed interval in X is compact.
Defn: f : (X, dx ) → (Y, dy ) is uniformly continuous if for all ǫ > 0, there exists a δ > 0 such that
dX (x1 , x2 ) < δ implies dY (f (x1 ), f (x2 )) < ǫ.
Pf hint: Let C = {y ∈ (a, b] | [a, y] can be covered
by a finite number of Uα } ∪ {a}. Let c = supC.
Defn: If A nonempty subset of metric space X and
x ∈ X, then the distance from x to A is
d(x, A) = inf {d(x, a) | a ∈ A}
Thm 27.3: A subspace of Rn is compact iff it is closed
and bounded in the Euclidean metric or the square
metric.
Idea of proof: (=>) compact Hausdorff implies closed.
For bounded, look at A ⊂ ∪∞
n=1 B(0, n)
Idea of proof: (<=) If A closed and bounded
A ⊂ B(0, r) ⊂ Π[−r, r]
Note: a set which is bounded in one metric can be
unbounded in a different metric even when both metrics generate the same topology.
Thm 27.4 (Extreme value thm).
f cont : (X, compact) → (Y, ordered) implies there
exists c, d ∈ X such that f (c) ≤ f (x) ≤ f (d) for all
x ∈ X.
Idea of proof: If f (X) has no largest element, then
f (X) ⊂ ∪y∈f (X) (−∞, y)
Note dA : X → [0, ∞), dA (x) = d(x, A) is a uniformly continuous function.
Idea of proof: Show that d(x, A) − d(y, A) ≤ d(x, y)
Defn: The diameter of A = diam(A) =
sup{d(a1 , a2 ) | a1 , a2 ∈ A}
Lemma 27.5 (Lebesgue number lemma)
Let X ⊂ ∪Uα . If X is compact, there is a δ > 0 such
that if diam(C) < δ, then there exists α0 such that
C ⊂ U α0
Idea of Proof: If X 6∈ {Uα | α ∈ A}, take a finite
subcover {Ui | i = 1, ..., n}. Let Ci = X − Ui .
Let f : X → R, f (x) = n1 Σd(x, Ci ). Note f is
continuous. Use extreme value thm to find m ∈ X
such that f (m) is the minimum value of f (X). Show
f (m) > 0 and let δ = f (m).
Thm 27.6 (Uniform continuity thm)
f cont : (X, compact metric) → (Y, metric) implies
f uniformly continuous.
−1
Idea of proof: X ⊂ ∪y∈Y f (BdY (y,
δ = Lebesgue number of this cover.
ǫ
2 ).
Let
A space, X, is compact if every open cover of X
contains a finite subcover.
A space, X, is compact if for every collection C of
closed sets in X having the finite intersection property, ∩C∈C C 6= ∅.
Defn: Given a topological space X, x is an isolated
point of X is {x} is open in X.
A space, X, is limit point compact if every infinite
subset of X has a limit point.
Thm 27.7: If X is a nonempty compact Hausdorff
space with no isolated points, then X is uncountable.
A space, X, is sequentially compact if every sequence
has a convergent subsequence.
Step 1: Take a nonempty open set U and take x ∈ X
(note x may or may not be in U ).
Thm 28.1 Compactness implies limit point compactness, but not conversely.
Use Hausdorff to find nonempty open set V such that
V ⊂ U and x 6∈ V .
Limit point compactness does not implies sequentially compactness (Hint: ((n, 1))∞
n=1 in Z+ ×{1, 2}).
Step 2: Suppose f : N → X, f (x) = xn . Show
f is not surjective (i.e., need to find a point not in
the image. Which definition of compact gives us a
point?).
Compactness does not imply sequential compactness
(Hint: [0, 1]ω ).
Thm 28.2: In a metrizable space X, TFAE:
1.) X is compact
2.) X is limit point compact
3.) X is sequentially compact.